kertrats wrote:I wrote this paper up last night. I'd like comments if anyone has any.

I like it :)

Very minor (hopefully constructive) criticisms:Some of your reference numbers seem to be off, likely due to the insertion of [2] or [3] after writing the rest of the paper.Equation(s) should probably be numbered too.

I was taught that table captions/labels should be positioned above the table, but I don't know if that's a widely accepted convention or just what my faculty wanted...

Of course, the problem with any equation is that the results are only as good as the data you put into it.

Based on studies, X_f should be 52 events/year and X_d averages 15 minutes if you use USA averages (not as optimistic as the e.g. numbers in the comic ). Use more regionally accurate numbers if available.

I'm looking at an orgy right now and it's WAY out of whack with the projected numbers. Maybe this is an anomaly but... what the hell am I posting for when I have these two girls 'all up 'pons'?? I don't think I'll even submit this.

Results for Ball State University, with 17000 students on campus at a time and with campus being ~1 mile and with the assumption that we are having sex 112 times a year (according to a Kinsley report...)

Right now, someone within 123.21 m of me is probably having sex. On the other hand, my roommate is definitely having sex within 2.1 m of me.

This comic finds the distance at which, on average, one couple within that distance from you is having sex. I want to know what the average distance from you to the nearest couple having sex is.

I will first verify that the comic's equation is correct, then take my best shot at answering my question.

First re-arrange the equation to read as follows: Xf*Xd*Pd*pi*r2 = 2Note that Xf*Xd is the probability that any one person is having sex.Note that Xf*Xd*Pd is the percentage (in decimal form) of people having sex in one square mile.Note that Xf*Xd*Pd*pi*r2 is the average number of people having sex in a circle of radius r.We set this last value equal to 2 because we assume it takes two to tango.We have now discovered that, on average, there will be one couple having sex within a circle of radius r.

So, the comic is correct. Now to answer my question:

Using statistical calculus, we find that the average radius among all the points in a circle of radius r is 2r/3.Therefore, the average distance from you to the nearest couple having sex is 2r/3.If r=150m, then this would be 100m.

How does that math look. Do others agree?I really can't figure out if that value is actually correct. Just because I found the average distance of the average area of, on average, one couple, doesn't mean that I found the average distance from you to the nearest couple having sex, does it?

kertrats wrote:I wrote this paper up last night. I'd like comments if anyone has any.

I think it's too Basically Decent that your Sexual Frequency table does not include an entry for <18 years old. Doubtless this is due to your sources. Perhaps you could try 4chan for the relevant data?

dln385 wrote:This comic finds the distance at which, on average, one couple within that distance from you is having sex. I want to know what the average distance from you to the nearest couple having sex is.

I will first verify that the comic's equation is correct, then take my best shot at answering my question.

First re-arrange the equation to read as follows: Xf*Xd*Pd*pi*r2 = 2Note that Xf*Xd is the probability that any one person is having sex.Note that Xf*Xd*Pd is the percentage (in decimal form) of people having sex in one square mile.Note that Xf*Xd*Pd*pi*r2 is the average number of people having sex in a circle of radius r.We set this last value equal to 2 because we assume it takes two to tango.We have now discovered that, on average, there will be one couple having sex within a circle of radius r.

So, the comic is correct. Now to answer my question:

Using statistical calculus, we find that the average radius among all the points in a circle of radius r is 2r/3.Therefore, the average distance from you to the nearest couple having sex is 2r/3.If r=150m, then this would be 100m.

How does that math look. Do others agree?I really can't figure out if that value is actually correct. Just because I found the average distance of the average area of, on average, one couple, doesn't mean that I found the average distance from you to the nearest couple having sex, does it?

If we assume one to tango there should be a 1 probability every computer connected to the Internet in a properly closed room. Which means the minimum radius is just the distance between you and that computer.

Using suspect data and noting that Vatican City has the seventh highest population density in the world (by country), I have calculated that on average, someone is having sex within about 400 meters of you, if you happen to be located within Vatican City.

I lament that I could not find data on the frequency of sex per year in Vatican City itself, and I am not at all convinced that 30 minutes is a very precise number, so my final result only has one significant digit. The distance would be reduced if Xd actually is more precise than I'm assuming. But what am I thinking. This is only an estimate.

Also, my calculation does not take into account the greater/lesser likelihood of sex in a country filled with priests.

If you ever attend a game in the Beaver Stadium which has a seating capacity of 107282 and an area of approximately 22500 square meters then, with the average American's rate of sex (about 2 - 3 per week so we'll say it's about 130 a year) and assuming that anyone having sex in a stadium will only do so for five minutes a time then someone is having sex within 10 meters of you

A more disturbing though is the radius at which, on average, someone is maturbating.

Guys guys guys! I found Russel's teapot! . . . nevermind, it was just Jesus flying to Mars again.

Well, to go with your reasoning yes, we are just estimating the probability, but until we see someone actually having sex they may or may not have it. However, I refute the applicability of quantum mechanics to sex. Have you ever tried going with a girl who is as wide as a few subatomic particles?

Well, to go with your reasoning yes, we are just estimating the probability, but until we see someone actually having sex they may or may not have it. However, I refute the applicability of quantum mechanics to sex. Have you ever tried going with a girl who is as wide as a few subatomic particles?

The other problem is that it tends to end up with the girl a little bit pregnant.

η

Ah but with quantum uncertainty, the girl isn't pregnant until you test

dln385 wrote:This comic finds the distance at which, on average, one couple within that distance from you is having sex. I want to know what the average distance from you to the nearest couple having sex is.

I will first verify that the comic's equation is correct, then take my best shot at answering my question.

First re-arrange the equation to read as follows: Xf*Xd*Pd*pi*r2 = 2Note that Xf*Xd is the probability that any one person is having sex.Note that Xf*Xd*Pd is the percentage (in decimal form) of people having sex in one square mile.Note that Xf*Xd*Pd*pi*r2 is the average number of people having sex in a circle of radius r.We set this last value equal to 2 because we assume it takes two to tango.We have now discovered that, on average, there will be one couple having sex within a circle of radius r.

So, the comic is correct. Now to answer my question:

Using statistical calculus, we find that the average radius among all the points in a circle of radius r is 2r/3.Therefore, the average distance from you to the nearest couple having sex is 2r/3.If r=150m, then this would be 100m.

How does that math look. Do others agree?I really can't figure out if that value is actually correct. Just because I found the average distance of the average area of, on average, one couple, doesn't mean that I found the average distance from you to the nearest couple having sex, does it?

For precise location, you have to take into account how energetic the particlesparticipants are.

I call Heisenberg on you.

Try the Printifier for xkcd.You can now scale the comic between 50 and 150%.

Am I the first to point out that the calculation is dead wrong?I'll present my case in logical order, but there's a short version in the end.

The 3rd dimension problem has been pointed out, so in the following I'm taking the formula to mean the horizontal distance.

1. Pd, the average number of people per area, is not useful, because people tend to aggregate.This is best seen in high residential buildings, where the average horizontal distance is much smaller than calculated.Taking this to the extreme, if everybody lived in high building (much higher than wide) which are very far-between (much much further-between than high),the actual distance would be much smaller than suggested by the low population density.

2. Let's look at the average fraction of time an average person is sexually active (it's not Xf*Xd, see next comment).The logic of the previous comment applies here too, because people tend to have sex at similar times.Suppose everybody had sex half of the time (OUCH! if you're a guy; WOW! if you're a girl), but only at night.The average distance would be very high, because half the time the closest lovemaking happens in a different time zone,much higher than suggested by all this much sex.

3. Xf and Xd could be correlated. Suppose sex lasted longer for people who were having it more frequently.In numbers, let's say half the people were having 45 minutes of sex every day, and the other half 15 minutes every 3 days.Then Xf*Xd would be 30mins*2/3days=1.389%,while the actual average fraction of time an average person was having sex would be (45mins*1/1day + 15mins*1/3days)/2 = 1.736%, which is 25% more.

4. Also, distance and frequency could be correlated. In order to calculate the average density of sexual activity, we can't just multiply Pd by Xf*Xd.Suppose people were evenly distributed in town, and everyone was having the same length sex,evenly spaced through the day, week and year (so that comments 1, 2 and 3 don't apply).Still, more sexually active people could be aggregated. If, for example, sex was mostly limited to a few far-between neighborhoods, we're back with the example of comment 1.

So:

We can't multiply Xf by Xd to get the average fraction of time spent having sex.Even if we could, we can't multiply that by Pd to get the average density of sexual activity.Even if we could, that would give us the average value for 2/pi*r^2 over all people and times.As comments 1 and 2 show, solving that for r does not give the average r over all people and times.

dln385 wrote:This comic finds the distance at which, on average, one couple within that distance from you is having sex. I want to know what the average distance from you to the nearest couple having sex is.

Actually, there is a slight subtelty here: The distance r in the comic is such that, on average, 1 person whithin this distance is having sex. This is what you prove. However, this is different from the average distance from you to the nearest person having sex.

And I would think it is more interesting to know the average minimum distance. Here is the formula I got: (making the same implicit dubious assumptions as in the comic)[math]d=\sqrt{\frac{1}{2P_d X_fX_d}}[/math]Hence [imath]d=\frac{\sqrt{\pi}}{2}r\simeq 0.887 r[/imath], and for instance r=150m in the comic gives d=133m for the average minimal distance.

How did I derive this formula? (for mathematically inclined) I assumed people (say, men) are distributed according to a two-dimensional Poisson point process with density [imath]\lambda=\frac{1}{2}P_dX_fX_d[/imath]. This is how "uniformly distributed" would translate in the most usual way. Then the number [imath]N_r[/imath] of people inside a circle of radius [imath]r[/imath] is a Poisson random variable with parameter [imath]\lambda\pi r^2[/imath], so that the distance [imath]D[/imath] to the nearest male having sex satisfies [imath]P(D>r)=P(N_r=0)=e^{-\lambda \pi r^2}[/imath]. We deduce the expected value [math]d=E[D]=\int_0^\infty P(D>r) dr=\int_0^\infty e^{-\lambda\pi r^2} dr=\frac{1}{2}\sqrt{\frac{1}{\lambda}}=\sqrt{\frac{1}{2P_dX_fX_d}}[/math]

I'd wondered about this, and thought that Randall's method of taking the city area per lovemaking occurrence would likely produce an inaccurate result. I was thinking some kind of 2D Poisson would be more applicable, but have only ever dealt with regular Poisson, so didn't know if this was possible. And now I find it is. Yay!

I had all kinds of plans in case of a zombie attack.I just figured I'd be on the other side.~ASW